The metal complex that is diamagnetic is: (Atomic number: Fe, 26; Cu, 29)

🧠 Diamagnetic = 0 Unpaired

Walk each:

🗺️ Test the Four

| Complex | Metal | OS | d-count | Unpaired | |---|---|---|---|---| | (1) $\mathrm{K_3[Cu(CN)_4]}$ | Cu | +1 | $\mathrm{d^{10}}$ | 0 ✓ | | (2) $\mathrm{K_2[Cu(CN)_4]}$ | Cu | +2 | $\mathrm{d^9}$ | 1 | | (3) $\mathrm{K_3[Fe(CN)_4]}$ | Fe | uncommon (would be Fe(–1) or so) | – | – | | (4) $\mathrm{K_4[FeCl_6]}$ | Fe | +2 | $\mathrm{d^6}$ + weak Cl⁻ → high-spin | 4 |

Only (1) $\mathrm{K_3[Cu(CN)_4]}$ has 0 unpaired (Cu is +1, $\mathrm{d^{10}}$).

OS check for (1): $3(+1) + x + 4(-1) = 0 \Rightarrow x = +1$. Cu(I), $\mathrm{d^{10}}$, fully paired → diamagnetic.

⚡ The "Cu(I) = Cu⁺ = d¹⁰ = Diamagnetic" Anchor

$\mathrm{Cu^+}$ ($\mathrm{d^{10}}$) is universally diamagnetic. Same goes for $\mathrm{Ag^+, Au^+, Hg^{2+}, Cd^{2+}, Zn^{2+}}$ — all d¹⁰. None of these can show paramagnetism in any geometry.

⚠️ Cu(I) vs Cu(II) From the Counter-Ion

The K count tells the OS:

• $\mathrm{K_3[Cu(CN)_4]}$ → 3K⁺ outside → complex charge $-3$ → Cu = $-3 - 4(-1) = +1$ → Cu(I).
• $\mathrm{K_2[Cu(CN)_4]}$ → 2K⁺ outside → complex $-2$ → Cu = $+2$ → Cu(II).

Always count K (or Na) to fix the OS.

$\boxed{\text{Answer: (1) } \mathrm{K_3[Cu(CN)_4]}}$